From a9dbd32bb13764399ee299cd17ef327f514d5673 Mon Sep 17 00:00:00 2001
From: Manoj Nayak <106598910+manojvn2612@users.noreply.github.com>
Date: Fri, 3 May 2024 13:01:39 +0530
Subject: [PATCH] Create furthest-building-you-can-reach.py

---
 Python/furthest-building-you-can-reach.py | 27 +++++++++++++++++++++++
 1 file changed, 27 insertions(+)
 create mode 100644 Python/furthest-building-you-can-reach.py

diff --git a/Python/furthest-building-you-can-reach.py b/Python/furthest-building-you-can-reach.py
new file mode 100644
index 00000000..ddbc9ea5
--- /dev/null
+++ b/Python/furthest-building-you-can-reach.py
@@ -0,0 +1,27 @@
+'''
+the condition was a person can use ladder or brick to climb a building but the sucessive building's height can be greater than or less than current buiiding.
+no. of bricks and ladder given,ladder length can be anything i.e if next buliding in 2 units greater than and we have no bricks so we can use ladder.
+so i used maxheap to know where should i use ladder so that it will not waste my resource and we can reach the farthest building.
+'''
+# import heapq
+class Solution:
+    def furthestBuilding(self, heights: list[int], bricks: int, ladders: int) -> int:
+        heap = []
+        for i in range(len(heights)-1):
+            diff = heights[i+1]-heights[i]
+            print(diff)
+            if diff<0:
+                continue
+            print("hello",diff)
+            bricks-=diff
+            print("b",bricks)
+            heapq.heappush(heap,diff)
+            print(heap)
+            if bricks<0:
+                if ladders==0:
+                    return i
+                ladders-=1
+                print("bricks",bricks,"ladder",ladders)
+                bricks+=heapq.heappop(heap)
+
+        return len(heights)-1